A $3$-bottom moldboard plow with a total width of cut of $90\text{ cm}$ operates at a depth of $15\text{ cm}$ and a forward speed of $5.4\text{ km/h}$. The specific soil draft resistance is $0.5\text{ kg/cm}^2$. Assuming the acceleration due to gravity $g = 9.81\text{ m/s}^2$, the drawbar power required to pull the plow is ________ $\text{kW}$ (round off to two decimal places).
Answer Key:
9.93
Step-by-Step Solution & Explanation
1. Cross-sectional area of furrow slice:
$A = \text{Width} \times \text{Depth} = 90\text{ cm} \times 15\text{ cm} = 1350\text{ cm}^2$
2. Total draft force required in kg:
$D_{\text{kg}} = A \times \text{Specific Resistance} = 1350\text{ cm}^2 \times 0.5\text{ kg/cm}^2 = 675\text{ kg}$
Converting to Newtons:
$D = 675 \times 9.81 = 6621.75\text{ N} = 6.622\text{ kN}$
3. Forward speed in m/s:
$v = 5.4\text{ km/h} = 5.4 \times \frac{5}{18} = 1.5\text{ m/s}$
4. Drawbar power:
$P_{\text{db}} = D \times v = 6.62175\text{ kN} \times 1.5\text{ m/s} = 9.9326\text{ kW} \approx 9.93\text{ kW}$